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Chapter 2: Problem 270

For the following exercises, point \(P\) and vector \(\mathbf{n}\) are given. a. Find the scalar equation of the plane that passes through \(P\) and has normal vector \(\mathbf{n}\). b. Find the general form of the equation of the plane that passes through \(P\) and has normal vector \(\mathbf{n}\). \(P(0,0,0), \quad \mathbf{n}=\langle-3,2,-1\rangle\)

Short Answer

Expert verified
a. Scalar equation: \(-3x + 2y - z = 0\). b. General form: \(-3x + 2y - z = 0\).

Step by step solution

01

Identify Given Values

We have a point \(P=(0,0,0)\) and a normal vector \(\mathbf{n}=\langle -3,2,-1 \rangle\). We will use these values to find the equation of the plane.
02

Use the Point-Normal Form

The point-normal form of the equation of a plane is: \(a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\), where \((x_0, y_0, z_0)\) are the coordinates of point \(P\) and \(a, b, c\) are the components of \(\mathbf{n}\).
03

Substitute and Simplify

Substitute \(a=-3\), \(b=2\), \(c=-1\), and \(P(0,0,0)\) into the point-normal form: \(-3(x-0) + 2(y-0) - 1(z-0) = 0\). Simplify to get: \(-3x + 2y - z = 0\).
04

Scalar Equation Derived

The scalar equation of the plane is obtained: \(-3x + 2y - z = 0\).
05

Rearrange to General Form

The general form of the equation of a plane is \(Ax + By + Cz + D = 0\). Our scalar equation is already in the general form: \(-3x + 2y - z = 0\), where \(A = -3\), \(B = 2\), \(C = -1\), and \(D = 0\).
06

Equation Verification

Verify that the given point \((0,0,0)\) satisfies the plane equation \(-3x + 2y - z = 0\). Substitute \(x=0\), \(y=0\), \(z=0\) into the equation: \(-3(0) + 2(0) - (0) = 0\). Therefore, the point lies on the plane.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Point-Normal Form
The point-normal form is an essential concept when dealing with the equations of planes. This form helps in deriving the equation of a plane from a point and a normal vector. The point-normal form of a plane's equation is given by:
  • \( a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \)
In this equation:
  • \( (x_0, y_0, z_0) \) are the coordinates of a point on the plane.
  • \( a, b, \text{ and } c \) represent the components of the normal vector.
For example, if you know a point \( P(0,0,0) \) and a normal vector \( \mathbf{n} = \langle -3, 2, -1 \rangle \), you substitute these into the point-normal form to start forming your plane equation.
Scalar Equation
Once you have the point-normal form, obtaining the scalar equation becomes straightforward. The scalar equation is essentially the simplified version of the point-normal form. By performing basic algebraic substitution and simplification, you derive the scalar equation.Start by substituting the point \( \(0,0,0\) \) and normal vector components \( \(a = -3, b = 2, c = -1\) \) into the point-normal form.
  • This yields: \[ -3(x-0) + 2(y-0) - 1(z-0) = 0 \]
  • Which simplifies to: \[ -3x + 2y - z = 0 \]
This equation is your scalar equation, a concise representation that defines the plane uniquely.
General Form of a Plane
The general form of a plane's equation is another useful format which is:
  • \( Ax + By + Cz + D = 0 \)
This format is especially helpful for analytical purposes, such as finding intersections or checking if points satisfy the plane equation.In this form:
  • \(A, B,\text{ and }C\) are the coefficients derived from the normal vector components, which in our case are \(A = -3, B = 2, C = -1\).
  • \(D\) is typically deduced from the constant terms in simplified equations, here \(D = 0\).
It's important to note that any valid scalar equation for a plane is already in the general form. Hence \( -3x + 2y - z = 0 \) perfectly matches the general form.
Normal Vector
The normal vector is a key element in defining a plane. It is a vector that is perpendicular to the plane's surface, and its components are critical in determining the plane's equation.In our example, the given normal vector is \( \mathbf{n} = \langle -3, 2, -1 \rangle \). Each component of this vector serves a purpose:
  • The first component, \(-3\), represents the influence on the \( x \) direction.
  • The second component, \(2\), relates to the \( y \) axis.
  • The third component, \(-1\), adjusts the \( z \) direction.
Understanding the normal vector makes it possible to construct the equation of the plane by aligning with point-normal, scalar, and general form methodologies.

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